拉格朗日插值用c语言怎么编程？各位高手帮帮忙啦

#include<stdio.h>

#include<string.h>

#define N 100

typedef struct tag{

double x

double y

}POINT

void main()

{

int i,j,n

double x,temp,Ln=0

POINT pt[N]

printf("请输入你要输入点的个数,,1<=n<=%d：\n",N)

printf("n=")

scanf("%d",&n)

printf("\n")

printf("\n请输入对应的点数\n")

for(i=0i<ni++)

scanf("%lf,%lf",&pt[i].x,&pt[i].y)

printf("\n")

printf("输入插值点x的值：\n")

scanf("%lf",&x)

printf("\n")

for(i=0i<ni++)

{

for(j=0,temp=1j<nj++)

{

if(j!=i)

temp=temp*(x-pt[j].x)/(pt[i].x-pt[j].x)

}

Ln=Ln+temp*pt[i].y

}

printf("输出：\nLn(%lf)=%lf\n",x,Ln)

}

我的程序是牛顿插值和拉格朗日插值合起来，你自己看下，用的是C++

#include <iostream>

#include <iomanip>

#include <stdlib.h>

using namespace std

#define N 100

void lagrange()

{

int n,k,m,q=1

float x[N],y[N],xx,yyy1,yyy2,yy1,yy2,yy3

cout<<"请输入X的个数："

cin>>n

for(k=0k<=n-1k++)

{

cout<<"请输入X"<<k<<"的值："

cin>>x[k]

cout<<"请输入Y"<<k<<"的值："

cin>>y[k]

}

system("cls")

cout<<"则Xi与Yi表格如下："<<endl

cout<<"Xi"<<" "for(k=0k<=n-1k++)cout<<setiosflags(ios::left)<<setw(10)<<x[k]

cout<<endl

cout<<"Yi"<<" "for(k=0k<=n-1k++)cout<<setiosflags(ios::left)<<setw(10)<<y[k]

cout<<endl

while(q)

{

cout<<"请输入所求x的值："

cin>>xx

while(xx>x[k-1]||xx<x[0])

{

cout<<"输入错误，请重新输入："

cin>>xx

}

for(k=0k<=n-1k++)

{

if(xx<x[k])

{

m=k-1

k=n-1

}

}

yyy1=y[m]*((xx-x[m+1])/(x[m]-x[m+1]))+y[m+1]*((xx-x[m])/(x[m+1]-x[m]))

cout<<"则拉格朗日分段线性插值为："<<yyy1<<endl

for(k=0k<=n-1k++)

{

if(xx<x[k])

{

m=k-1

k=n-1

}

}

if((xx-x[m])>(x[m+1]-xx))m=m+1

else m=m

yy1=y[m-1]*((xx-x[m])*(xx-x[m+1]))/((x[m-1]-x[m])*(x[m-1]-x[m+1]))

yy2=y[m]*((xx-x[m-1])*(xx-x[m+1]))/((x[m]-x[m-1])*(x[m]-x[m+1]))

yy3=y[m+1]*((xx-x[m-1])*(xx-x[m]))/((x[m+1]-x[m-1])*(x[m+1]-x[m]))

yyy2=yy1+yy2+yy3

cout<<"则拉格朗日分段二次插值为："<<yyy2<<endl

cout<<"是否输入其余要求x的值[是(1)，否(0)]："

cin>>q

}

system("cls")

}

void main()

{

lagrange()

}

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